Question

A simple random sample of 100 bags of tortilla chips produced by Company X is selected every hour for quality control. In the current sample, 18 bags had more chips (measured in weight) than the labeled quantity. The quality control inspector wishes to use this information to calculate a 90% confidence interval for the true proportion of bags of tortilla chips that contain more than the label states.
1. What is the value of the standard error of the sample proportion?
a. 0.038
b. 384
c. 0.0015
d. 0.063

Answer 1

Answer:

Option A) 0.038

Step-by-step explanation:

We are given the following in the question:

Sample size, n = 100

Number of of bags of tortilla chips that contain more than the label states, x = 18

Sample proportion:

$\hat{p} = \dfrac{x}{n} = \dfrac{18}{100} = 0.18$

Formula for standard of error:

$\sqrt{\dfrac{\hat{p}(1-\hat{p})}{n}}$

Putting the values, we get:

$\sqrt{\dfrac{0.18(1-0.18)}{100}} =0.038$

Thus, the correct answer is

Option A) 0.038